Posted by **Melody** on Thursday, October 15, 2009 at 2:51am.

I have another question I am stuck on that follows a previous question. How do I get an equation for the height of the car on the ferris wheel over a 10 minute period if we start the analysis one minute into the ride.

Can I still use the original formula

h = -35cos(2pi*X/5)+37

I don't understand what part changes to make it one minute into the ride. Appreciate any help, orignal question below.

have a trig question regarding a ferris wheel, radius 35m, axles 37m, clockwise direction, rotates twice every 10 mins. I have calculated the following function:

h = -35cos(2pi*X/5)+37

I have calculated at what height I would be after 3 minutes by substituting X=3 in the above equation. However, I need to calculate at what times I would be 30m off the ground. Can you please tell me if I am on the right track and how to calculate the time?

Many thanks

Maths - Reiny, Tuesday, October 13, 2009 at 8:28am

I agree with your equation, so let' set it equal to 30

30 = -35cos(2pi*X/5)+37

-7 = -35cos(2pi*X/5)

.2 = cos(2pi*X/5)

take the inverse cosine to get

2pi*X/5 = 1.369438

x = 1.09 minutes

check it by repeating the same steps you did finding the height at 3 minutes.

- Math -
**Reiny**, Thursday, October 15, 2009 at 11:12am
I am not too sure what you mean by "if we start the analysis one minute into the ride"

Where do you want the chair to be at 1 minute. The way it is, the chair would be 26.2 m high, (which also confirms my answer of 30 m at 1.09 minutes)

Do you want the chair to be at a minimum height of 2 m when the time is 1 minute?

If that is the case, we will have to perform a phase shift on our equation.

So we want h = 2 at t=1, rather than t=0

this will do it:

h = -35cos(2pi/5)(X-1) + 37

check: remember it takes 5 minutes for one rotation.

t = 1, h = 2 , at the bottom

t = 2.25 , h = 37, the height of the axle, check!

t = 3.5, h = 72 , the height of axle + radius of wheel, check!

t = 4.75, h = 37 , coming down, height of axle, check!

t = 6 , h = 2, back at the bottom, check!

In the original equation, using t values of 0, 1.25, 2.5, 3.75 and 5 will yield the same heights.

(That is how I checked if your equation was valid)

BTW, I gave you a time of 1.09 minutes for a height of 30 m.

That would be on the way up, of course you would reach another height of 30 m on your way down, which would be at (5-1.09) or 3.91 min.

repeatedly adding 5 minutes to these answers would produce all the times that the chair is 30 m high.

t =

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